If D is a point on the side AB of ∆ABC such that AD : DB = 3.2 and E is a point on BC such that DE || AC.

Solution:

Given: In ΔABC, D is a point on side AB such that AD : DB= 3 : 2. E is a point on side BC such that DE || AC.

To find: $\frac{\Delta \mathrm{ABC}}{\Delta \mathrm{BDE}}$

In ΔABC,

$\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{3}{2}$

Since DE\|AC,

$\frac{E C}{E B}=\frac{3}{2}$

According to converse of basic proportionality theorem if a line divides any two sides of a triangle in the same ratio, then the line must be parallel to the third side.

Hence DE || AC

In ΔBDE and ΔABC,

∠BDE=∠A             Corresponding angles

∠DBE=∠ABC       Common

So, ∆BDE~∆ABC      (AA Similarity)

We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides.

Let AD = 2x and BD = 3x.

Hence

$\frac{A r(\triangle \mathrm{ABC})}{A r(\triangle \mathrm{BDE})}=\frac{\mathrm{AB}^{2}}{\mathrm{BD}^{2}}$

$=\frac{(\mathrm{BD}+\mathrm{DA})^{2}}{(\mathrm{BD})^{2}}$

$=\frac{(3 x+2 x)^{2}}{(2 x)^{2}}$

$=\frac{(5 x)^{2}}{(2 x)^{2}}$

$\frac{A r(\Delta \mathrm{ABC})}{\operatorname{Ar}(\Delta \mathrm{BDE})}=\frac{25}{4}$